JEE Main 2021 · 27 July · evening · Single correct · +4 / −1
The resistance of a conductor at 15°C is 16Ω and at 100°C is 20Ω. What will be the temperature coefficient of resistance of the conductor?
- (a) 0.010°C⁻¹
- (b) 0.033°C⁻¹
- (c) 0.003°C⁻¹
- (d) 0.042°C⁻¹
Answer and solution
Answer (c) 0.003°C⁻¹
Solution
Physics sir · 4 steps
Write R = R₀(1 + α t) at both temperatures and divide, so R₀ drops out and only α is left.
- Set up. Resistance grows linearly with temperature: Rₜ = R₀(1 + α t), where R₀ is the resistance at 0 °C. Write this once at 15 °C and once at 100 °C.
R₁₀₀/R₁₅ = (1 + 100α)/(1 + 15α) = 20/16
- Solve for alpha. Cross-multiply: 16 + 1600α = 20 + 300α, so 1300α = 4. This gives α = 4/1300 ≈ 3.1 × 10⁻³ °C⁻¹.
- Quick check. The shortcut α ≈ (Δ R)/(R₁ Δ t) = 4/(16 × 85) ≈ 2.9 × 10⁻³ °C⁻¹ gives the same rounded value.
- So. The temperature coefficient is about 0.003 °C⁻¹, which is option (c).
The idea to keep. When you get resistance at two temperatures, take the ratio of R₀(1 + α t) at each one. The unknown R₀ cancels and you can solve for α in one line.
Answer from the NTA final answer key. Solution by PrepWiser.